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Difficulty: MediumRadical and Rational Equations

What is the sum of all real values of xx that satisfy the equation 3x+40−x=4\sqrt{3x + 40} - x = 4?

  1. A
    -5
  2. 3Answer
  3. C
    5
  4. D
    -8

Answer

The sum of all valid real solutions is 3.
Isolating the radical gives 3x+40=x+4\sqrt{3x + 40} = x + 4. Squaring both sides and simplifying results in x2+5x−24=0x^2 + 5x - 24 = 0. Factoring gives (x+8)(x−3)=0(x + 8)(x - 3) = 0, leading to candidate solutions of 3 and -8. Substituting these back shows that only 3 satisfies the original equation, as -8 results in an extraneous solution. Therefore, the sum of all valid solutions is 3.

Step-by-Step Solution

1
Isolate the radical expression on one side of the equation.
3x+40=x+4\sqrt{3x + 40} = x + 4
To prepare both sides of the equation for squaring.
2
Square both sides of the equation to eliminate the radical.
3x+40=(x+4)23x + 40 = (x + 4)^2
Squaring a square root cancels the radical.
3
Expand the right side and move all terms to one side to form a quadratic equation.
x2+5x−24=0x^2 + 5x - 24 = 0
Expanding (x+4)2(x+4)^2 yields x2+8x+16x^2 + 8x + 16. Subtracting 3x3x and 4040 from both sides sets the quadratic equation to zero.
4
Factor the quadratic equation.
(x+8)(x−3)=0(x + 8)(x - 3) = 0
The factors of -24 that add up to 5 are 8 and -3.
5
Solve for the potential values of xx.
x=−8x = -8 or x=3x = 3
Setting each factor to zero yields the candidate solutions.
6
Check each candidate solution in the original equation to identify any extraneous solutions.
For x=3x = 3, 3(3)+40−3=7−3=4\sqrt{3(3)+40}-3 = 7-3 = 4 (valid). For x=−8x = -8, 3(−8)+40−(−8)=4+8=12≠4\sqrt{3(-8)+40}-(-8) = 4+8 = 12 \neq 4 (extraneous).
Squaring both sides can introduce extraneous solutions that do not satisfy the original equation.

Key Concept

Solving radical equations and checking for extraneous solutions.
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Topics

Radical and Rational Equations

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